Is ${896067}$ divisible by $3$ ?
Answer: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {896067}= &&{8}\cdot100000+ \\&&{9}\cdot10000+ \\&&{6}\cdot1000+ \\&&{0}\cdot100+ \\&&{6}\cdot10+ \\&&{7}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {896067}= &&{8}(99999+1)+ \\&&{9}(9999+1)+ \\&&{6}(999+1)+ \\&&{0}(99+1)+ \\&&{6}(9+1)+ \\&&{7} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {896067}= &&\gray{8\cdot99999}+ \\&&\gray{9\cdot9999}+ \\&&\gray{6\cdot999}+ \\&&\gray{0\cdot99}+ \\&&\gray{6\cdot9}+ \\&& {8}+{9}+{6}+{0}+{6}+{7} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${896067}$ is divisible by $3$ if ${ 8}+{9}+{6}+{0}+{6}+{7}$ is divisible by $3$ Add the digits of ${896067}$ $ {8}+{9}+{6}+{0}+{6}+{7} = {36} $ If ${36}$ is divisible by $3$ , then ${896067}$ must also be divisible by $3$ Add the digits of ${36}$ $ {3}+{6} = \color{#9D38BD}{9} $ If $\color{#9D38BD}{9}$ is divisible by $3$ , then ${36}$ must also be divisible by $3$ $\color{#9D38BD}{9}$ is divisible by $3$, therefore ${896067}$ must also be divisible by $3$.